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1499. Max Value of Equation

Description

You are given an array points containing the coordinates of points on a 2D plane, sorted by the x-values, where points[i] = [xi, yi] such that xi < xj for all 1 <= i < j <= points.length. You are also given an integer k.

Return the maximum value of the equation yi + yj + |xi - xj| where |xi - xj| <= k and 1 <= i < j <= points.length.

It is guaranteed that there exists at least one pair of points that satisfy the constraint |xi - xj| <= k.

 

Example 1:

Input: points = [[1,3],[2,0],[5,10],[6,-10]], k = 1
Output: 4
Explanation: The first two points satisfy the condition |xi - xj| <= 1 and if we calculate the equation we get 3 + 0 + |1 - 2| = 4. Third and fourth points also satisfy the condition and give a value of 10 + -10 + |5 - 6| = 1.
No other pairs satisfy the condition, so we return the max of 4 and 1.

Example 2:

Input: points = [[0,0],[3,0],[9,2]], k = 3
Output: 3
Explanation: Only the first two points have an absolute difference of 3 or less in the x-values, and give the value of 0 + 0 + |0 - 3| = 3.

 

Constraints:

  • 2 <= points.length <= 105
  • points[i].length == 2
  • -108 <= xi, yi <= 108
  • 0 <= k <= 2 * 108
  • xi < xj for all 1 <= i < j <= points.length
  • xi form a strictly increasing sequence.

Solutions

Solution 1

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class Solution:
    def findMaxValueOfEquation(self, points: List[List[int]], k: int) -> int:
        ans = -inf
        pq = []
        for x, y in points:
            while pq and x - pq[0][1] > k:
                heappop(pq)
            if pq:
                ans = max(ans, x + y - pq[0][0])
            heappush(pq, (x - y, x))
        return ans

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class Solution {
    public int findMaxValueOfEquation(int[][] points, int k) {
        int ans = -(1 << 30);
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[0] - a[0]);
        for (var p : points) {
            int x = p[0], y = p[1];
            while (!pq.isEmpty() && x - pq.peek()[1] > k) {
                pq.poll();
            }
            if (!pq.isEmpty()) {
                ans = Math.max(ans, x + y + pq.peek()[0]);
            }
            pq.offer(new int[] {y - x, x});
        }
        return ans;
    }
}

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class Solution {
public:
    int findMaxValueOfEquation(vector<vector<int>>& points, int k) {
        int ans = -(1 << 30);
        priority_queue<pair<int, int>> pq;
        for (auto& p : points) {
            int x = p[0], y = p[1];
            while (pq.size() && x - pq.top().second > k) {
                pq.pop();
            }
            if (pq.size()) {
                ans = max(ans, x + y + pq.top().first);
            }
            pq.emplace(y - x, x);
        }
        return ans;
    }
};

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func findMaxValueOfEquation(points [][]int, k int) int {
    ans := -(1 << 30)
    hp := hp{}
    for _, p := range points {
        x, y := p[0], p[1]
        for hp.Len() > 0 && x-hp[0].x > k {
            heap.Pop(&hp)
        }
        if hp.Len() > 0 {
            ans = max(ans, x+y+hp[0].v)
        }
        heap.Push(&hp, pair{y - x, x})
    }
    return ans
}

type pair struct{ v, x int }

type hp []pair

func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool {
    a, b := h[i], h[j]
    return a.v > b